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Re: x and y are positive integers. If the greatest common divisor of 3x
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02 Apr 2017, 06:09
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GMATPrepNow wrote:
x and y are positive integers. If the greatest common divisor of 3x and 3y is 6, what is the value of y?
(1) The greatest common divisor of 2x and 2y is 2y (2) The least common multiple of 2x and 2y is 20
Target question:What is the value of y?
Given: The greatest common divisor of 3x and 3y is 6 This means that, if we examine the prime factorization of 3x and prime factorization of 3y, they will share exactly ONE 3 and ONE 2. That is: 3x = (2)(3)(?)(?)(?)(?) 3y = (2)(3)(?)(?)(?)(?) NOTE: Both prime factorizations might include other primes, BUT there is no additional overlap beyond the ONE 3 and ONE 2.
Notice that if we divide both sides of both prime factorizations by 3, we get: x = (2)(?)(?)(?)(?) y = (2)(?)(?)(?)(?)
This tells us that the greatest common divisor (GCD) of x and y is 2.
Statement 1: The greatest common divisor of 2x and 2y is 2y We already know that... x = (2)(?)(?)(?)(?) y = (2)(?)(?)(?)(?)
So, 2x = (2)(2)(?)(?)(?)(?) 2y = (2)(2)(?)(?)(?)(?) This tells us that the greatest common divisor (GCD) of 2x and 2y =(2)(2) = 4.
The statement tells us that the GCD of 2x and 2y is 2y, which means 2y = 4 Solve the equation to get y = 2. PERFECT!! Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: The least common multiple of 2x and 2y is 20 There are several values of x and y that satisfy statement 2 (as well as satisfying the given information). Here are two: Case a: x = 2 and y = 10. In this case, 2x = 4 and 2y = 20, and the least common multiple of 4 and 20 is 20. Also notice that 3x = 6 and 3y = 30, and the GCD of 6 and 30 is 6, which satisfies the given information. In this case y = 10 Case b: x = 10 and y = 2. In this case, 2x = 20 and 2y = 4, and the least common multiple of 20 and 4 is 20. Also notice that 3x = 30 and 3y = 6, and the GCD of 30 and 6 is 6, which satisfies the given information. In this case y = 2 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
x and y are positive integers. If the greatest common divisor of 3x
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02 Apr 2017, 05:49
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GMATPrepNow wrote:
x and y are positive integers. If the greatest common divisor of 3x and 3y is 6, what is the value of y?
(1) The greatest common divisor of 2x and 2y is 2y (2) The least common multiple of 2x and 2y is 20
*kudos for all correct solutions
Again a very good Q..
x and y are integers and GCD of 3x and 3y is 6.. This means x and y have only one common factor that is 6/3=2..
Let's see the statements
(1) The greatest common divisor of 2x and 2y is 2y this tells us that x is multiple of y... But we already know that ONLY common factor between x and y is 2, so y will be 2 Sufficient
(2) The least common multiple of 2x and 2y is 20 LCM of x and y would be 10.. But both x and y are multiples of 2.. So x and y could be 2 and 10, OR 10 and 2 So y can be 2 or 10 Insufficient
x and y are positive integers. If the greatest common divisor of 3x
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02 Apr 2017, 15:40
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Great question as always.
x and y are positive integers. If the greatest common divisor of 3x and 3y is 6, what is the value of y?
Analyzing the prompt:
GCD of 3x and 3y is 6.......it means that both numbers MUST have 2 in its prime factorization, with one of them with power of 1. By definition GCD consists of prime factors with smallest exponent. For example:
3x=12...x=2^2 & 3y=6...y=2...............GCD=6
3x=6.....x=2 & 3y=12..y=2^2..........GCD=6
3x=12...x=2^2 & 3y=18..y=6..............GCD=6
3x=12...x=2^2 & 3y=30...y=10...........GCD=6 (Notice that 30 has 5 so smallest factor is 5^0 in 12)
(1) The greatest common divisor of 2x and 2y is 2y
Spotting the statement above means that 2x is dividable by 2y.
2x/2y= 2y.......x=2y^2....... it becomes the following:
GCD of 6y^2 & 3y = 6.......There is one solution which is y=2.
Re: x and y are positive integers. If the greatest common divisor of 3x
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11 May 2018, 03:24
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GCF(3x,3y) = 6 => implies that x = 2m, y = 2n, where m and n are co-primes
Statement 1: GCF(2x, 2y) = 2y => which means y is multiple of x, and this can happen only when y = 2 (n = 1). sufficient Statement 2: LCM(2x,2y) = 20 I can see two cases, x = 2, y = 2 * 5 or x = 2 * 5 , y = 2 not sufficient